Energy Balance

PROCESS CONDITIONS

Solid Components at 303.15 K

Total Mass Flow Rate 10987.18592 kg/day

Mass of BaSO4 6776.159157 kg/day

Mass of ZnO 50.07507506 kg/day

Mass of ZnS 3188.78078 kg/day

Inlet Temperature (T1) 50 0C 323.15 K

Outlet Temperature (T2) 125 0C 398.15 K

Air at 523.15 K

Mass Flow Rate 1269266300 kg/day

Constant Pressure Heat Capacity 1.0407388 kJ/kg-K

Inlet Temperature (t1) 515 0C 788.15 K

Outlet Temperature (t2) 335 0C 608.15 K

For the calculation of wet bulb temperature, for spray dryers, Nt is in general between 0.25 to 1. (Land,

2012). Assume the value of Nt =1

= ln (

)

1 = ln (788.15 608.15

)

= 503.3941928

Calculating the logarithmic mean temperature difference:

=( ) ( )

ln (

)

=(788.15 503.3942) (608.15 503.3942)

ln ((788.15 503.3942)608.15 503.3942 )

= 180

CALCULATION OF HEAT DUTY

Calculation of Heat Duty:

i. Sensible heat required to bring the solids to the final temperature:

Assume: Solids before entering the dryer are at 500C:

Solving for the calorific temperature of the solid components:

=1 + 2

2=

323.15 + 398.15

2

= 360.65

Using Perrys Handbook Cp solid components @ 360.65 K: (Perry & Green, 2008)

Cp of BaSO4 6.144707242 kJ/kg-K

Cp of ZnO 2.445455874 kJ/kg-K

Cp of ZnS 2.709484455 kJ/kg-K

1 = ( )

1 = (6776.159157

) (6.144707242

kJ

kg K) (398.15 323.15)

+ (50.07507506

) (2.445455874

kJ

kg K) (398.15 323.15)

+ (3188.78078

) (2.709484455

kJ

kg K) (398.15 323.15)

3717841.093

= 154910.0455

ii. Sensible heat to raise the temperature of water from initial (30 C) to vaporization temperature (50C

323.15 K)

Using Perrys Handbook, Cp of liquid water

T Cp

323.15 K 4.185257778 kJ/kg K

2 = ( )

2 = (972.1709048

) (4.185257778

kJ

kg K) (373.15 323.15)

2 = 203439.292

= 8476.637168

iii. Latent heat to vaporize the moisture:

Using Perrys Handbook, latent heat of vaporization at 373.15 K (1000C):

T Latent heat

373.15 K 2258.234 kJ/kg

3 = ().

3 = (971.6848193

) (2258.234

kJ

kg)

3 = 2194291

= 91428.80268

iv. Sensible heat to raise temperature of water from vaporization temperature to final product temperature

(1250C):

Solving for the calorific temperature of solid component:

=1 + 2

2=

373.15 + 398.15

2= 385.65 = 112.50

Using Perrys Handbook, Cp of liquid water

T Cp

385.25 K 4.235077 kJ/kg K

4 = ( ).

4 = (972.1709048 971.6848193)(4.235077)(398.15 373.15)K

4 = 51.46523735

= 2.144384889

Total heat requirement of the system:

= 1 + 2 + 3 + 4

= (3717841 + 203439 + 2194292 + 51.4652) = 6115623.547

Using Ambient Temperature of 26 C and RH = 89% (Yesterday's and last week's weather in manila, 2015)

Using Psychometric Chart from Perrys Chemical Engineers Handbook:

Humidity = 0.01901 kg of water per kg of dry air

Taking the humid heat of Air:

= 1.005 + 1.88

= 1.005 + 1.88(0.01901)

= 1.0407388

Calculating for the amount of air needed.

=

=

6115623.547

1.0407388

(788.15 608.15)

= 1057721917

For the humidity of outlet air:

=971.6848193

1057721917+ 0.01901

kg water

kg air

= 00.019010919 kg water

kg air

Allotting an allowance of 20%, for the possible heat losses and due to startup, shutdown and cleaning:

(Land, 2012)

= 1.20 (1057721917

)

= 1269266300